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+//Example 6.20

+clc

+disp("V_CC = 20 V,  R_L = 20 ohm, turns ratio 1.58:1")

+n=1/1.58

+format(7)

+disp(n,"  n = 1/1.58 = ")

+rl=20/0.6329^2

+disp(rl,"Therefore,  R''_L(in ohm) = R_L / n^2 =")

+disp("(i) For maximum possible peak to peak output voltage, the power output is also maximum possible. For this condition the slope of the a.c. load line can be expressed as")

+disp("R''_L = V_m/I_m = V_CC/I_CQ")

+icq=20/49.928

+format(4)

+disp(icq,"Therefore,  I_CQ(in A) =")

+ibq=0.4/40

+format(5)

+disp(ibq,"Therefore,  I_BQ(in A) = I_CQ/beta =")

+disp("This is the required value of the base current")

+disp("(ii)  P_ac = I_Irms^2 * R''_L")

+disp("But for maximum power output condition,")

+irms=0.4/sqrt(2)

+format(8)

+disp(irms,"I_Irms(in A) = I_Im/sqrt(2) = I_CQ/sqrt(2) =")

+pac=49.928*0.2828^2

+format(2)

+disp(pac,"Therefore,  P_ac(in W) =")

+disp("(iii)  %eta = P_ac/P_DC * 100")

+pdc=20*0.4

+disp(pdc,"Now  P_DC(in W) = V_CC * I_CQ =")

+eta=400/8

+format(3)

+disp(eta,"%eta(in percentage) =")